I am studying Newtonian Mechanics and I am familiar with single variable calculus.

I came across the concept of conservative and non conservative forces and potential energy. Here is what I understand:

A force is said to be conservative if the work done by the force on an object/system is irrelevant of the path taken, and only depends on the initial and final position of the point of the application of the force. Also, for conservative forces, work done on a closed path is zero. For example, gravity.

A force like friction which does not satisfy the above property, it is said to be a non conservative force.

I also understand that any system cannot possess any potential energy if only non conservative forces act on it as it would be dissipated in form of internal and dissipative forces. Thus, when we are talking about potential energy associated with a system we are talking about conservative forces.

But what i cannot understand is why $f_x = -\frac{du}{dx}$ ; $f_y = -\frac{du}{dy}$ ; $f_z = -\frac{du}{dz}$

Can someone please prove this to me, as well as explain the intuition behind it?

I read a proof online which showed that the work done by the gravity on a ball is negative hence $dw = -du$ from which it follows $f = -\frac{du}{dy}$ I understood this proof, but doesn’t this prove that the above result is valid only in the case of gravitational potential energy, and that too only when the object is moving upwards? Am i missing something from this?

Also please explain the intuition behind the force potential energy relation.

  • $\begingroup$ Hi Srinidhi kabra. Welcome to Phys.SE. Are you asking about the minus? $\endgroup$
    – Qmechanic
    Jun 25 '21 at 14:37
  • $\begingroup$ Yes Sir. I cannot understand why there should be a minus sign. $\endgroup$ Jun 25 '21 at 14:45
  • $\begingroup$ More on sign conventions and potential energy. $\endgroup$
    – Qmechanic
    Jun 25 '21 at 15:40
  • $\begingroup$ Thank you very much for you help kind Sir. $\endgroup$ Jun 25 '21 at 16:36

The potential energy function is a definition. For any function $U(x,y,z)$ whose gradient$^*$ $\nabla U$ exists we can look at the relation $-\nabla U=\mathbf F$ and say that $\mathbf F$ is a conservative force and $U$ is is corresponding potential energy.

In physics we tend to go the other way: given a force $\mathbf F$, if we can find a function $U$ such that $-\nabla U=\mathbf F$, then $U$ is the corresponding potential energy function. Typically you can find $U$ through integration. But even then, it's still just a definition.

This is helpful because, via the fundamental theorem of calculus, the work done by a conservative force is path-independent. For path $C$ with end points $\mathbf a$ and $\mathbf b$ we can find the work done by the conservative force by only looking at the end points of the path:

$$W=\int_C\mathbf F\cdot\text d\mathbf x=\int_C-\nabla U\cdot\text d\mathbf x=U(\mathbf a)-U(\mathbf b)=-\Delta U$$

$^*$The gradient is just shorthand for the derivatives you are asking about: it's a vector of derivatives $$\nabla U=\frac{\partial U}{\partial x}\,\hat x+\frac{\partial U}{\partial y}\,\hat y+\frac{\partial U}{\partial z}\,\hat z$$

  • 1
    $\begingroup$ Just to clarify, so the potential energy function is defined that way on purpose ? If so, then what is the physical intuition behind it ? $\endgroup$ Jun 25 '21 at 14:44
  • $\begingroup$ @Srinidhikabra It's a helpful definition since you can determine the work done by conservative forces easily via the potential energy function regardless of the path taken. For non-conservative forces the work is path-dependent. $\endgroup$ Jun 25 '21 at 14:47
  • 2
    $\begingroup$ Would you want mathematics to have defined the derivative of a single valued funciton $df/dx = - \lim_{\Delta x \to 0} \Delta f/\Delta x$? with a minus sign? If not, why would you want the gradient of a multivalued function to have a built in minus sign? The minus sign comes from the physics, not from the math. $\endgroup$
    – alephzero
    Jun 25 '21 at 15:01
  • $\begingroup$ @alephzero I don't think I talk about multivalued functions in my answer. As for the negative sign, I agree, it is very useful for the physics. $\endgroup$ Jun 25 '21 at 15:40
  • $\begingroup$ @GiorgioP Ugh yep. thanks! $\endgroup$ Jun 25 '21 at 15:50

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